Maths Qs. Pg 40 Text Test 6

iCrazzy

Taken from Q.20 AND 19

Q.20
The curve y=ax^2+bx+c contains the points (0,1)(2,9)(4,41)
Using these points write 2 simultaneous equations in a b and c
hence solve the equation to find the values of a b and c.
Now I subbed them all in and got
this:
a+b+c=1 -A
4a+2b+c=9 -B
16a+4b+c=41 -C
but when I solve the equations using the 3 variables of simultaneous the answer does not work out to what it says on the back of the book.
first I solve a+b
then I solve a+c
and using the answer to both I solve them together but I get wrong, so if someone could point me where I am getting wrong

q.19 is
A farmer builds a long narrow pen for sheep using 60m of fencing.
if he doubles the width and halves the length he only needs to use 42m of fencing.
find the dimensions of the two pens.

Now what I did was:
x+y=60
4x+y=84
but I am sure that is wrong...

Please if someone could could help me with these two would be much appreciated.

Moody_mona

Your first equation of the simultaneous is wrong, try subbing in 0 for x again...

To make a pen, a farmer will need two lengths and two widths, use that sentence to make the first equation and go from there.

Blue giant

iCrazzy wrote: »

Taken from Q.20 AND 19

Q.20
The curve y=ax^2+bx+c contains the points (0,1)(2,9)(4,41)
Using these points write 2 simultaneous equations in a b and c
hence solve the equation to find the values of a b and c.
Now I subbed them all in and got
this:
a+b+c=1 -A
4a+2b+c=9 -B
16a+4b+c=41 -C
but when I solve the equations using the 3 variables of simultaneous the answer does not work out to what it says on the back of the book.
first I solve a+b
then I solve a+c
and using the answer to both I solve them together but I get wrong, so if someone could point me where I am getting wrong

q.19 is
A farmer builds a long narrow pen for sheep using 60m of fencing.
if he doubles the width and halves the length he only needs to use 42m of fencing.
find the dimensions of the two pens.

Now what I did was:
x+y=60
4x+y=84
but I am sure that is wrong...

Please if someone could could help me with these two would be much appreciated.

Well for Q20 firstly the first equation should read c=1. Anything multiplied by 0 is 0. Should work out then?

Can't see anything wrong with Q19. Did you try and work it out? X=8 and Y=52 I think ( don't have any paper or pen ).
Edit: What he said ^^^^

iCrazzy

Moody_mona wrote: »

Your first equation of the simultaneous is wrong, try subbing in 0 for x again...

You mean this :
a+b+c=1 -A

so would that mean even you you multiply 0 by letters it still is zero ?

Blue giant

iCrazzy wrote: »

You mean this :
a+b+c=1 -A

so would that mean even you you multiply 0 by letters it still is zero ?

Yes. Anything by zero is zero.

iCrazzy

Blue giant wrote: »

Yes. Anything by zero is zero.

Then how am I supposed to solve these:
c=1 -A
4a+2b+c=9 -B
16a+4b+c=41 -C

I would solve b+c no problem but how would I solve A with B or C.
would it be normal like this:
41+2b+c=9
c=`
4a+2b+2c=10

Blue giant

Sub in c=1 in all of the equations and then it's a simple simultaneous equation with two variables. You've been doing them since 1st year. A=3 B=-4 I think ?

Moody_mona

You now know what c is, so you can sub c=1 into your second two equations.

iCrazzy

Moody_mona wrote: »

You now know what c is, so you can sub c=1 into your second two equations.

Thank you.

Take Your Pants Off

Hey lads, any help in this. I have an upcoming maths test on monday and I am dreading it.
For Q.12 and 13.
In Q13 I only got as far as -8p squared - 4q squared < 0

DublinArnie

Take Your Pants Off wrote: »

Hey lads, any help in this. I have an upcoming maths test on monday and I am dreading it.
For Q.12 and 13.
In Q13 I only got as far as -8p squared - 4q squared < 0

Working on it.

DublinArnie

Q13 is a complex number so it has no roots. The question asked to prove it, you got it right. I just went on to further factorise it for fun, but I think you got it right because it asked TO show.

wmirl

Take Your Pants Off wrote: »

Hey lads, any help in this. I have an upcoming maths test on monday and I am dreading it.
For Q.12 and 13.
In Q13 I only got as far as -8p squared - 4q squared < 0

Q13 is incorrectly phrased. [latex]p = q = 0[/latex] results in a repeated real solution of [latex]x=0[/latex]. It's trivial but still a solution, the question should have excluded p, q both being zero.

You're correct in having [latex] b^2 -4ac = -8p^2 - 4q^2 = -4(2p^2 + q^2)[/latex]. The terms inside the brackets are positive since p, q are both real, their squares are positive or zero and multiplying by -4 results in [latex]b^2 - 4ac \le 0 [/latex], which corresponds to the complex root case or the repeated roots ([latex]p=q=0[/latex]), but if the question stated that p, q were not both zero, the complex root would be the only case, i.e. no real roots.

DublinArnie wrote: »

Q13 is a complex number so it has no roots. The question asked to prove it, you got it right. I just went on to further factorise it for fun, but I think you got it right because it asked TO show.

You've chosen the wrong b in your workings, you've included a constant term, instead of only the coefficient of x.

Peregrine

As for the second part of Q12, once you have expressed a in terms of b, all you have to do is use the quadratic formula to find the root.

[latex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/latex]

But [latex]b^2 - 4ac = 0[/latex] and [latex] a = \frac{b^2}{4}[/latex]

So,

[latex]x = \frac{-b \pm \sqrt{0}}{2(\frac{b^2}{4})}[/latex]

You can finish off from there.